Respuesta :
Answer:
See the steps below.
Step-by-step explanation:
a. Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries.
The sample means = 0.17
The standard deviation of the sample means = [tex]\sqrt{0.17*(0.83/800)}[/tex] = 0.0133
Therefore, the sample proportion of households spending more than $100 per week on groceries is approximately 0.0133.
b) What is the probability that the sample proportion will be within plus or minus .02 of the population proportion?
Since the sample means = 0.17
For plus 0.02
The sample means = 0.17 + 0.02 = 0.19
Therefore, we have:
z(0.19) = [tex](0.19 - 0.17)/\sqrt{0.17*(0.83/800)}[/tex] = 1.5060
For minus 0.02
The sample means = 0.17 - 0.02 = 0.15
z(0.15) = [tex](0.15 - 0.17)/\sqrt{0.17*(0.83/800) }[/tex] = - 1.5060
Therefore, we now have:
P(0.15< p < 0.19) = P(-1.5060< z <1.5060) = 0.8679
c) Answer part (b) for a sample of 1600.
For plus 0.02
The sample means = 0.17 + 0.02 = 0.19
Therefore, we have:
z(0.19) = [tex](0.19 - 0.17)/\sqrt{0.17*(0.83/1,600)}[/tex] = 2.1297
For minus 0.02
The sample means = 0.17 - 0.02 = 0.15
z(0.15) =[tex](0.15 - 0.17)/\sqrt{0.17*(0.83/1,600)}[/tex] = - 2.1297
Therefore, we now have:
P(0.15< p < 0.19) = P(-2.197< z <2.1297) = 0.9660
Answer:
A)sample proportion = 0.17, the sampling distribution of p can be calculated/approximated with normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281
B) 0.869
C)0.9668
Step-by-step explanation:
A) p ( proportion of population that spends more than $100 per week) = 0.17
sample size (n)= 800
the sample proportion of p = 0.17
standard error of p = [tex]\sqrt{\frac{p(1-p)}{n} }[/tex] = 0.013281
the sampling distribution of p can be calculated/approximated with
normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281
B) probability that the sample proportion will be +-0.02 of the population proportion
= p (0.17 - 0.02 ≤ P ≤ 0.17 + 0.02 ) = p( 0.15 ≤ P ≤ 0.19)
z value corresponding to P
Z = [tex]\frac{P - p}{standard deviation}[/tex]
at P = 0.15
Z = (0.15 - 0.17) / 0.013281 = = -1.51
at P = 0.19
z = ( 0.19 - 0.17) / 0.013281 = 1.51
therefore the required probability will be
p( -1.5 ≤ z ≤ 1.5 ) = p(z ≤ 1.51 ) - p(z ≤ -1.51 )
= 0.9345 - 0.0655 = 0.869
C) for a sample (n ) = 1600
standard deviation/ error = 0.009391 (applying the equation for calculating standard error as seen in part A above)
therefore the required probability after applying
z = [tex]\frac{P-p}{standard deviation}[/tex] at p = 0.15 and p = 0.19
p ( -2.13 ≤ z ≤ 2.13 ) = p( z ≤ 2.13 ) - p( z ≤ -2.13 )
= 0.9834 - 0.0166 = 0.9668
